- 3 - 



Putting p = f(t) - — '. — J_i-(t) 

 3 Ti- 



pR, 



(7) becomes 



^rom wh ich 



♦ i I / + 



3 77 



ky = f(t) 



y = _ f f(u) sin q(t-u) du 

 Mq J» 



for the conditions y = y = o at t = -0D 

 where 



M = m + 



2 k 



^ = R 



37T 



297 



(8) 



(9) 



(10) 



Ap plt cation to Subtle ^'-'avcs . 



The bubb)e wave from 1 oz. charges may be represented approximately by 

 p(t) = p^exp (-^" ) 



(n> 



Numerical evaluation of the last term in equation (u) using equation (ll) for conditions 

 typical of R.R.L. box model tests showed that its value was about the same as the equivalent term for 

 the pressure at the centre point; viz. p 



i'-'r) 



The solutions of (lO) are: 

 for finite baffle, e quat ion (4) , taking centre point value for the last term in (t) 



exp(nt) 



-={!iriS)J 



n^. q2 



2 exp(-nt) - exi 



"(-?). 



2 2 



n + q 



In sin qt 

 n +q q 



for - a. < t < o 



for < t < **2 



>(12) 



- ^xp !!i\ 



exp(-nt) 



n^. q^ 



q(h^+q^) 



J 2 sin qt - sin q A - ^ \l for ^ < t < < 



for finite baffle, equation (5j 

 My exp{nt) 



2x 2 



for - 00 < t < o 



for < t < -JO 



= exp(-nt) ^ 2n sin qt 

 n *q q(n +q ) 



for the infinite baffle, equation (6), the deflection is twice that in equation (l3). 



(13) 



Piston 



